package org.lep.leetcode.distinctsubsequences;

import java.util.Arrays;

/**
 *
 * Source : https://oj.leetcode.com/problems/distinct-subsequences/
 *
 * Created by lverpeng on 2017/8/17.
 *
 * Given a string S and a string T, count the number of distinct subsequences of T in S.
 *
 * A subsequence of a string is a new string which is formed from the original string
 * by deleting some (can be none) of the characters without disturbing the relative positions
 * of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
 *
 * Here is an example:
 * S = "rabbbit", T = "rabbit"
 *
 * Return 3.
 */
public class DistinctSubsequences {

    /**
     *
     * 求解的个数，使用动态规划
     *
     * 状态：
     * i,j表示T中长度为i的prefix:T[0:i-1]，S中长度为j的prefix:S[0-j-1],S[j]第j个字符，T[i]第i个字符
     * DP[i][j]表示：S[0:j]中包含T[0:i]唯一子串的个数，当j<i的时候DP[i][j] = 0
     *
     * 递推公式：
     * 当S[j] != T[i]的时候
     * DP[i+1][j+1] = DP[i+1][j]，含义是当前字符不相等的时候，S[j+1]包含T[i+1]的个数就是S[j]包含T[i+1]的个数
     *
     * 当S[j] == T[i]的时候
     * DP[i+1][j+1] = DP[i+1][j] + DP[i][j]，含义是当前字符相等的时候，S[j+1]包含T[i+1]的个数就是S[j]包含T[i+1]的个数加上S[j]包含T[i]的个数
     *
     * 计算方向和起始状态：
     * DP[i][j]
     * DP[i+1][j],DP[i+1[j+1]
     * 从上到下，从左到右
     *
     * 第0行：1
     * 第0列：0
     *
     *
     * @param S
     * @param T
     * @return
     */
    public int distinctSequences (String S, String T) {
        int[][] dp = new int[T.length()+1][S.length()+1];
        for (int i = 0; i <= T.length(); i++) {
            dp[i][0] = 0;
        }

        for (int i = 0; i <= S.length(); i++) {
            dp[0][i] = 1;
        }
        for (int i = 1; i <= T.length(); i++) {
            for (int j = i; j <= S.length(); j++) {
                if (S.charAt(j-1) == T.charAt(i-1)) {
                    dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
                } else {
                    dp[i][j] = dp[i][j-1];
                }
            }
        }
        return dp[T.length()][S.length()];
    }


    /**
     * 优化DP占用空间，因为递推的时候只需要dp[i][j-1],dp[i-1][j-1]
     * 也就是当前矩阵左上角的值和左面的值，使用滚动数组优化空间
     *
     * @param S
     * @param T
     * @return
     */
    public int distinctSequences1 (String S, String T) {
        int[] dp = new int[S.length()+1];
        Arrays.fill(dp, 1);
        for (int i = 1; i <= T.length(); i++) {
            int upLeft = dp[0];
            dp[0] = 0;
            for (int j = 1; j <= S.length(); j++) {
                // 相当于记下dp[i-1][j-1]
                int temp = dp[j];
                // 相当于dp[i][j-1]
                dp[j] = dp[j-1];
                if (S.charAt(j-1) == T.charAt(i-1)) {
                    dp[j] += upLeft;
                }
                upLeft = temp;
            }
        }
        return dp[S.length()];
    }

    public static void main(String[] args) {
        DistinctSubsequences subsequences = new DistinctSubsequences();
        System.out.println(subsequences.distinctSequences("rabbbit", "rabbit") + "------3");
        System.out.println(subsequences.distinctSequences1("rabbbit", "rabbit") + "------3");
    }
}
